/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 Suppose the mapping \(F: \mathbf... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose the mapping \(F: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}\) is defined by \(F(x, y)=(x+y, x) .\) Show that \(F\) is linear.

Short Answer

Expert verified
To show that the mapping F is linear, we need to prove it satisfies scalar multiplication and vector addition conditions. For scalar multiplication, we find that \(F(ax, ay) = (ax + ay, ax)\) and \(aF(x,y) = (a(x+y), ax)\), and since they are equal, F satisfies the scalar multiplication condition. For vector addition, we find that \(F(x1 + x2, y1 + y2) = (x1 + x2 + y1 + y2, x1 + x2)\) and \(F(x1, y1) + F(x2, y2) = (x1 + y1 + x2 + y2, x1 + x2)\), and since they are equal, F satisfies the vector addition condition. Therefore, F is linear.

Step by step solution

01

Check Scalar Multiplication Condition

Let (x, y) be a vector in $$R^2$$, and let a be a scalar in $$R$$. The definition of scalar multiplication is given by: $$F(a(x, y)) = F(ax, ay).$$ Recall that the mapping function F is defined by: $$F(x, y) = (x + y, x).$$ Therefore, we can substitute the values of ax and ay into the mapping function: $$F(ax, ay) = (ax + ay, ax).$$ We can also determine the scalar multiplication F(av) for vector (x, y) as: $$aF(x,y) = a(x+y, x) = (a(x+y), ax).$$
02

Compare Scalar Multiplication Expressions

Now, we can compare F(ax, ay) to aF(x, y) to see if they are equal: $$F(ax, ay) = (ax + ay, ax),$$ $$aF(x,y) = (a(x+y), ax).$$ Since these two expressions are equal, the mapping F satisfies the scalar multiplication condition.
03

Check Vector Addition Condition

Let (x1, y1) and (x2, y2) be two vectors in $$R^2$$. The definition of vector addition is given by: $$F((x1, y1) + (x2, y2)) = F(x1+x2, y1+y2).$$ Now, let's substitute the values of (x1+x2) and (y1+y2) into the mapping function: $$F(x1 + x2, y1 + y2) = (x1 + x2 + y1 + y2, x1 + x2).$$ We can also determine the vector addition F(v) + F(w) for vectors (x1, y1) and (x2, y2): $$F(x1, y1) + F(x2, y2) = (x1 + y1, x1) + (x2 + y2, x2).$$
04

Compare Vector Addition Expressions

Now, we can compare F(x1 + x2, y1 + y2) to F(x1,y1) + F(x2,y2) to see if they are equal: $$F(x1 + x2, y1 + y2) = (x1 + x2 + y1 + y2, x1 + x2),$$ $$F(x1, y1) + F(x2, y2) = (x1 + y1 + x2 + y2, x1 + x2).$$ Since these two expressions are equal, the mapping F satisfies the vector addition condition. Since F satisfies both the scalar multiplication and vector addition conditions, we can conclude that F is linear.

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Most popular questions from this chapter

Label the following statements as true or false. (a) Every linear operator on an \(n\)-dimensional vector space has \(n\) distinct eigenvalues. (b) If a real matrix has one eigenvector, then it has an infinite number of eigenvectors. (c) There exists a square matrix with no eigenvectors. (d) Eigenvalues must be nonzero scalars. (e) Any two eigenvectors are linearly independent. (f) The sum of two eigenvalues of a linear operator \(T\) is also an eigenvalue of \(T\). (g) Linear operators on infinite-dimensional vector spaces never have eigenvalues. (h) An \(n \times n\) matrix \(A\) with entries from a field \(F\) is similar to a diagonal matrix if and only if there is a basis for \(\mathrm{F}^{n}\) consisting of eigenvectors of \(A\). (i) Similar matrices always have the same eigenvalues. (j) Similar matrices always have the same eigenvectors. (k) The sum of two eigenvectors of an operator \(\mathrm{T}\) is always an eigenvector of \(T\).

Consider the mapping \(F: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2}\) defined by \(F(x, y, z)=\left(y z, x^{2}\right) .\) Find (a) \(F(2,3,4)\) (b) \(F(5,-2,7)\) (c) \(F^{-1}(0,0)\), that is, all \(v \in \mathbf{R}^{3}\) such that \(F(v)=0\)

Determine whether or not each of the following linear maps is nonsingular. If not, find a nonzero vector \(v\) whose image is 0 (a) \(F: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}\) defined by \(F(x, y)=(x-y, x-2 y)\) (b) \(G: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}\) defined by \(G(x, y)=(2 x-4 y, 3 x-6 y)\)

Suppose that \(F: V \rightarrow U\) is linear and that \(V\) is of finite dimension. Show that \(V\) and the image of \(F\) have the same dimension if and only if \(F\) is nonsingular. Determine all nonsingular linear mappings \(T: \mathbf{R}^{4} \rightarrow \mathbf{R}^{3}.\)

Suppose \(V\) has finite dimension. Suppose \(T\) is a linear operator on \(V\) such that \(\operatorname{rank}\left(T^{2}\right)=\operatorname{rank}(T) .\) Show that \(\operatorname{Ker} T \cap \operatorname{Im} T=\\{0\\}\)

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